Upper Bound for Lucas Number

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Theorem

Let $L_n$ denote the $n$th Lucas number.

Then:

$L_n < \paren {\dfrac 7 4}^n$


Proof

The proof proceeds by complete induction.

For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:

$L_n < \paren {\dfrac 7 4}^n$


$\map P 1$ is the case:

\(\ds L_1\) \(=\) \(\ds 1\)
\(\ds \) \(<\) \(\ds \dfrac 7 4\)

Thus $\map P 1$ is seen to hold.


Basis for the Induction

$\map P 2$ is the case:

\(\ds L_2\) \(=\) \(\ds 3\)
\(\ds \) \(=\) \(\ds \dfrac {48} {16}\)
\(\ds \) \(<\) \(\ds \dfrac {49} {16}\)
\(\ds \) \(=\) \(\ds \paren {\dfrac 7 4}^2\)

Thus $\map P 2$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.


This is the induction hypothesis:

$L_k < \paren {\dfrac 7 4}^k$


from which it is to be shown that:

$L_{k + 1} < \paren {\dfrac 7 4}^{k + 1}$


Induction Step

This is the induction step:

\(\ds L_{k + 1}\) \(=\) \(\ds L_k + L_{k - 1}\)
\(\ds \) \(<\) \(\ds \paren {\dfrac 7 4}^k + \paren {\dfrac 7 4}^{k - 1}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {\dfrac 7 4}^{k - 1} \paren {1 + \dfrac 7 4}\)
\(\ds \) \(=\) \(\ds \paren {\dfrac 7 4}^{k - 1} \paren {\dfrac {11} 4}\)
\(\ds \) \(<\) \(\ds \paren {\dfrac 7 4}^{k - 1} \paren {\dfrac 7 4}^2\)
\(\ds \) \(=\) \(\ds \paren {\dfrac 7 4}^{k + 1}\)

So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{\ge 1}: L_n < \paren {\dfrac 7 4}^n$

$\blacksquare$


Sources

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